I have learnt combinations and I have been attempting at these two questions but couldn't solve it:

1) In a mixed pack of coloured light bulbs there are three red bulbs, one yellow bulb, one blue bulb and one green bulb. Four bulbs are selected at random from the pack. How many different selections are possible?

I did 3C1 * 1C1 * 1C1 * 1C1 = 3

However, that wasn't the answer.

2) There are 20 teachers at a conference. Of these, 8 are maths teachers, 6 are history teachers, 4 are physics teachers and 2 are geography teachers.

Four of the teachers are to be chosen at random to take part in a quiz.

In how many different ways can the teachers be chosen if there are to be at least two maths teachers?

So I did 8C2 * 18C2 = 4284

Which also wasn't the answer. Could someone please tell me what I am doing wrong?

Thanks in advance.


1 Answer 1


As regards the first problem, divide it into pairwise disjoint cases where you select $k=1,2,3$ red bulbs. Then, by using the "rule of sum", the number of ways is $$\sum_{k=1}^3 \binom{3}{4-k}=1+3+3=7.$$ Here you can also enumerate the seven cases explicitly: RBGY ($k=1$), RRBG, RRBY, RRGY ($k=2$), RRRB, RRRG, RRRY ($k=3$).

For the second one, consider the complement case when you have $0$ math teachers, $m_0=\binom{20-8}{4}$ ways, and $1$ math teacher, $m_1=?$ ways. Then the number of ways such that teachers can be chosen with at least two maths teachers is $$\binom{20}{4}-m_0-m_1=\binom{20}{4}-\binom{20-8}{4}-?.$$ Can you take it from here?

  • $\begingroup$ If you don't mind, could you please explain why I wouldn't use the rule of multiplication in this case? $\endgroup$
    – ianc1339
    Sep 14, 2017 at 7:28
  • $\begingroup$ No I mean I thought I was supposed to use the rule of multiplication but you are instead adding. So could you please explain why you would add $\endgroup$
    – ianc1339
    Sep 14, 2017 at 7:37
  • $\begingroup$ @kimchiboy03 I am using the "rule of sum" because i'm considering pairwise disjoint cases en.wikipedia.org/wiki/Rule_of_sum $\endgroup$
    – Robert Z
    Sep 14, 2017 at 7:39
  • $\begingroup$ @kimchiboy03 See my edited answer. Are you able to evaluate $m_0$ and $m_1$? $\endgroup$
    – Robert Z
    Sep 14, 2017 at 7:40
  • $\begingroup$ Would m0 be 8C0 and m1 be 8C1? $\endgroup$
    – ianc1339
    Sep 14, 2017 at 7:46

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